Question
Step by step explanation hybridization of water
Answers
GPT-4o mini
The hybridization of water (H₂O) can be understood by examining the molecular orbital theory along with the concept of atomic orbitals from the contributing atoms (oxygen and hydrogen). Here’s a step-by-step explanation:
### Step 1: Identify the Central Atom
In a water molecule, the central atom is oxygen (O). The molecular formula for water is H₂O, indicating that it has two hydrogen atoms bonded to one oxygen atom.
### Step 2: Determine the Electron Configuration of Oxygen
Oxygen has an atomic number of 8, which means it has 8 electrons. The electron configuration of oxygen is:
\[ \text{1s}^2 \text{2s}^2 \text{2p}^4 \]
In the second shell, oxygen has:
- 2 electrons in the 2s orbital
- 4 electrons in the 2p orbitals (which can hold a total of 6 electrons)
### Step 3: Count the Valence Electrons
Oxygen has 6 valence electrons (2 from the 2s orbital and 4 from the 2p orbitals). Each hydrogen atom has 1 valence electron, contributing a total of 2 from the two hydrogen atoms. Therefore, the total number of valence electrons in water is:
\[ 6 \text{(from O)} + 2 \text{(from 2 H)} = 8 \text{ valence electrons} \]
### Step 4: Determine the Required Hybridization
To form the two O-H bonds in water, oxygen needs to hybridize its orbitals. The bonding in water involves two single bonds (sigma bonds) formed between the oxygen and hydrogen atoms.
### Step 5: Hybridization Process
1. **Promotion of Electrons**: The oxygen atom uses its hybridization to make bonds. In the case of water, one of the 2s electrons is promoted to an empty 2p orbital, allowing the oxygen atom to effectively have four orbitals to work with (two 1s orbitals and three combined orbitals).
2. **Hybridization**: Oxygen undergoes sp³ hybridization. The 2s orbital combines with two 2p orbitals to form four equivalent sp³ hybrid orbitals. This hybridization generates four orbitals, which are arranged in a tetrahedral geometry.
### Step 6: Bond Formation
In water, two of the sp³ hybrid orbitals form sigma bonds with hydrogen atoms, while the other two hybrid orbitals contain lone pairs of electrons. The arrangement of the two hydrogen atoms and the two lone pairs leads to a bent molecular geometry, with a bond angle of about 104.5 degrees.
### Step 7: Geometry and Polarity
The bent shape of the water molecule and the presence of oxygen's electronegativity leads to a polar molecule. The oxygen atom has a partial negative charge (δ-) while the hydrogen atoms have a partial positive charge (δ+).
### Summary
- Oxygen hybridizes from its 2s and some 2p orbitals to form four sp³ hybrid orbitals.
- Two of the sp³ hybrid orbitals bond with hydrogen atoms, and the other two hold lone pairs of electrons, resulting in a bent structure for the water molecule.
- The overall hybridization of water is sp³, which contributes to its unique properties, like being a polar solvent.
By understanding these steps, you can grasp the hybridization concept and the bonding in water molecules.
### Step 1: Identify the Central Atom
In a water molecule, the central atom is oxygen (O). The molecular formula for water is H₂O, indicating that it has two hydrogen atoms bonded to one oxygen atom.
### Step 2: Determine the Electron Configuration of Oxygen
Oxygen has an atomic number of 8, which means it has 8 electrons. The electron configuration of oxygen is:
\[ \text{1s}^2 \text{2s}^2 \text{2p}^4 \]
In the second shell, oxygen has:
- 2 electrons in the 2s orbital
- 4 electrons in the 2p orbitals (which can hold a total of 6 electrons)
### Step 3: Count the Valence Electrons
Oxygen has 6 valence electrons (2 from the 2s orbital and 4 from the 2p orbitals). Each hydrogen atom has 1 valence electron, contributing a total of 2 from the two hydrogen atoms. Therefore, the total number of valence electrons in water is:
\[ 6 \text{(from O)} + 2 \text{(from 2 H)} = 8 \text{ valence electrons} \]
### Step 4: Determine the Required Hybridization
To form the two O-H bonds in water, oxygen needs to hybridize its orbitals. The bonding in water involves two single bonds (sigma bonds) formed between the oxygen and hydrogen atoms.
### Step 5: Hybridization Process
1. **Promotion of Electrons**: The oxygen atom uses its hybridization to make bonds. In the case of water, one of the 2s electrons is promoted to an empty 2p orbital, allowing the oxygen atom to effectively have four orbitals to work with (two 1s orbitals and three combined orbitals).
2. **Hybridization**: Oxygen undergoes sp³ hybridization. The 2s orbital combines with two 2p orbitals to form four equivalent sp³ hybrid orbitals. This hybridization generates four orbitals, which are arranged in a tetrahedral geometry.
### Step 6: Bond Formation
In water, two of the sp³ hybrid orbitals form sigma bonds with hydrogen atoms, while the other two hybrid orbitals contain lone pairs of electrons. The arrangement of the two hydrogen atoms and the two lone pairs leads to a bent molecular geometry, with a bond angle of about 104.5 degrees.
### Step 7: Geometry and Polarity
The bent shape of the water molecule and the presence of oxygen's electronegativity leads to a polar molecule. The oxygen atom has a partial negative charge (δ-) while the hydrogen atoms have a partial positive charge (δ+).
### Summary
- Oxygen hybridizes from its 2s and some 2p orbitals to form four sp³ hybrid orbitals.
- Two of the sp³ hybrid orbitals bond with hydrogen atoms, and the other two hold lone pairs of electrons, resulting in a bent structure for the water molecule.
- The overall hybridization of water is sp³, which contributes to its unique properties, like being a polar solvent.
By understanding these steps, you can grasp the hybridization concept and the bonding in water molecules.